Chapter 13: Q. 43 (page 1027)

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in $ℝ^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
$\int_{0}^{2 π} \int_{0}^{3} (6 r - 2 r^{2}) d r d θ$

Short Answer

Expert verified

The value of integral is

$\int_{0}^{2 π} \int_{0}^{3} (6 r - 2 r^{2}) d r d θ = 18 π$

Step by step solution

Given information

The expression is

I = \int_{0}^{2 π} \int_{0}^{3} (6 r - 2 r^{2}) d r d θ

Calculation

Here, $r = 0, r = 3$ and $θ = 0, θ = 2 π$

Integrate with respect to $r$ first,

$I = \int_{0}^{2 π} {[\frac{6 r^{2}}{2} - \frac{2 r^{3}}{3}]}_{0}^{3} d θ [\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C]$

Put the limits

localid="1651390685586" $I = \int_{0}^{2 π} [\frac{6 (3)^{2}}{2} - \frac{2 (3)^{3}}{3}] d θ I = \int_{0}^{2 π} 9 d θ$

Now integrate with respect to $θ$

$I = 9 [θ]_{0}^{2 π} I = 18 π$

Thus, the value of integral is

$\int_{0}^{2 π} \int_{0}^{3} (6 r - 2 r^{2}) d r d θ = 18 π$

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Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in $ℝ^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
$\int_{0}^{2 π} \int_{0}^{3} (6 r - 2 r^{2}) d r d θ$

Short Answer

Step by step solution

Given information

Calculation

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Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in ℝ3. Use polar coordinates to describe the solid, and evaluate the expressions.∫02π∫036r-2r2drdθ

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in $ℝ^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
$\int_{0}^{2 π} \int_{0}^{3} (6 r - 2 r^{2}) d r d θ$